In probability problems why do 'and' and 'or' covert to '$\times$' and '+'? Title says it all - would anyone kindly explain why 'and' translates as '$\times$' rather than '+'?
 A: Consider the following examples


*

*OR: You are in corssroad and you can follow 3 directions (left, right or straight). You can choose each with probability $1/3$. What is the probability that you will choose left or straight? Obviously $1/3+1/3=2/3$.

*And: A school has $40\%$ boys and $50\%$ of boys like soccer. What is the probability that a randomly chosen kid is a boy and likes soccer? Obviously the boys that like soccer are the $20\%$ of the total population of kids in the school, since they are the $50%$ of the $40%$ which is equal to $50\%\cdot40\%=20\%$. This is the required probability.
The intuition behind this examples is that OR means more choises so you add up probabilities, thus $+$, but AND means more restrictive characterization (you have to be that and that...), thus $\cdot$. Note, to this, that multiplying two nonnegative numbers that are less than $1$ you receive an even "smaller" number (still nonnegative and less than the other two).
A: Because of the properties of measure and the definition of independence.
If you have 2 disjoint events $A$ and $B$, then the probability of $A$ or $B$ is $\mu(A\cup B) = \mu(A)+\mu(B)$.
If you have 2 independent events, they satisfy (if $\mu(A)\neq0\neq\mu(B)$)
$$\frac{\mu(B\cap A)}{\mu(A)}=\mu(B)\Longleftrightarrow\frac{\mu(A\cap B)}{\mu(B)}=\mu(A)$$
And from this it's obvious that the probability of $A$ and $B$ is $\mu(A\cap B)=\mu(A)\cdot\mu(B)$.
A: Truth is usually denoted by $1$, while false is usually $0$. Now see the "and" relation:
$$T\wedge F=F,$$
or, equivalently,
$$1\wedge 0=0,$$
which reminds of multiplication:
$$1\cdot 0=0.$$
On the other hand, see the "or" relation:
$$T\vee F=T,$$
or, equivalently,
$$1\vee0=1,$$
which reminds of addition:
$$1+0=1.$$
This is why conjunction is usually called "logical multiplication" and disjunction is "logical addition".
